Linear System with Parameter
Here is problem 3 from the Fifth International Internet Mathematical Olympiad for Students. This is an online competition run by the Ariel University Center of Samaria, Israel.
Find all values of the parameter y for which the following system has a solution
(1)  x_{5} + x_{2}  = yx_{1} 
(2)  x_{1} + x_{3}  = yx_{2} 
(3)  x_{2} + x_{4}  = yx_{3} 
(4)  x_{3} + x_{5}  = yx_{4} 
(5)  x_{4} + x_{1}  = yx_{5} 
For each value of y, find all the solutions.
Contact Front page Contents Algebra Store
Copyright © 19962017 Alexander Bogomolny
Find all values of the parameter y for which the following system has a solution
(1)  x_{5} + x_{2}  = yx_{1} 
(2)  x_{1} + x_{3}  = yx_{2} 
(3)  x_{2} + x_{4}  = yx_{3} 
(4)  x_{3} + x_{5}  = yx_{4} 
(5)  x_{4} + x_{1}  = yx_{5} 
For each value of y, find all the solutions.
First of all observe that, for any value of y, the system has a trivial solution:
(y  2)(x_{1} + x_{2} + x_{3} + x_{4} + x_{5}) = 0. 
So, either y = 2 or x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 0. Clearly the value
But, to follow up, assume first that
(1')  x_{5} + x_{2}  = 2x_{1} 
(2')  x_{1} + x_{3}  = 2x_{2} 
(3')  x_{2} + x_{4}  = 2x_{3} 
(4')  x_{3} + x_{5}  = 2x_{4} 
(5')  x_{4} + x_{1}  = 2x_{5} 
Since the system is homogeneous and we are looking for a nontrivial solution, we may omit the last equation and solve the rest with the Gaussian elimination, getting subsequently,
 ⇒ 

By the back substitution, x_{5} = x_{4} = x_{3} = x_{2} = x_{1}, with arbitrary x_{1}.
Now we may assume y ≠ 2 and x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 0. We'll replace (quite arbitrarily) the last equation with this one and employ again the Gaussian elimination.
 ⇒ 
 ⇒ 
 ⇒ 
 ⇒  
 ⇒ 

If y² + y  1 ≠ 0, the last equation gives x_{5} = 0 and the back substitution leads to a trivial solution which was already accounted for. So let us assume that
y_{1,2} = (1 ± √5) / 2. 
With the parameter y equal to one of them, the last equation implies that x_{5} can be chosen arbitrarily; and, from the fourth equation, the same is true of x_{4}.
Since, for the chosen y, y(y + 1) = 1, the third equation gives
x_{3} = yx_{4}  x_{5} 
From the second equation,
x_{2} = y(x_{4} + x_{5}) 
and, from the first equation,
x_{1} = x_{4} + yx_{5}. 
Thus the system has three sets of solutions:
A trivial solution for any y,
A 1parameter solution for y = 2,
A 2parameter solution for y = (1 ± √5) / 2.
The latter family, perhaps surprisingly, includes the golden ratio.
Contact Front page Contents Algebra Store
Copyright © 19962017 Alexander Bogomolny
62100713 